Print the environment variables WITHOUT VALUES in C - c

How to print the environment variables in C, but WITHOUT VALUES ?? Only variables.
int main(int argc, char **argv, char **envp)
{
while(*envp!=NULL) {
printf("%s\n", *envp);
envp++;
}
system("pause");
return 0;
}

Since environmental variables have format of NAME=value you need to display only part of string up to = character.
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **argv, char **envp)
{
while(*envp!=NULL) {
char * len = strchr(*envp, '=');
if (len == NULL)
printf("%s\n", *envp);
else
printf("%.*s\n", len - *envp, *envp);
envp++;
}
system("pause");
return 0;
}
Ideone

Environment variables are of the form NAME=value. So, you can look for the first = sign and print only upto it to get only the names.

Related

I want to split a string into two strings in C

I want to split a string by the comma and separate the first number in the string into its own new string, the rest of the string I want to keep together.
So far I have tried this by using strtok() and I can get the first number into its own string, but now I can't figure out how to keep the rest of the string together.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>
int main(int argc, char *argv[])
{
char testStr[] = "1000,first,second,third,+abc";
char *uidStr;
char *restOfstr;
int n;
//This is wrong, I know, but I want to populate
//the rest of the string after the first comma
//into a single string without the UID.
uidStr = strtok(testStr, ",");
while (n < 5)
{
restOfstr = strtok(NULL, ",");
n++;
}
return 0;
}
strtok works fine, you have to keep in mind that it returns a pointer to each tokenized word so you need two pointers one for the first token and other for the rest of the string.
Demo
#include <stdio.h>
#include <string.h>
int main()
{
char testStr[] = "1000,first,second,third,+abc";
char *uidStr; //pointer to uid
char *restOfstr; //pointers to the rest of the string
uidStr = strtok(testStr, ","); //uid remains in testStr
restOfstr = strtok(NULL, "\n"); //rest of the string
puts(uidStr); //or puts(testStr) to print uid
puts(restOfstr); //print rest of the string
return 0;
}
If you want a more secure function you can use strtok_s.
You can use strchr to find the first comma in the string.
Then using strncpy to get the number in the string.
The complete code:
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
char *str = "1000,first,second,third,+abc";
char *s = strchr(str, ',');
if(!s)
return -1;
char num[10];
strncpy(num, str, s-str);
num[s-str] = '\0';
int a = strtol(num, NULL, 10);
printf("num = %d\nthe remaining: %s\n", a, s+1);
return 0;
}
#include <string.h>
#include <stdio.h>
int main(int ac, char **av) {
while (--ac) {
char *p = *++av;
char *t = strtok(p, ",");
char *r = strtok(NULL,"");
printf("%s : %s\n", t, r);
}
return 0;
}
Note that the empty string "" passed to the second strtok means that it cannot find a deliminator, thus returns the rest of the string.
In addition to the excellent answers #mevets and #anastaciu have provided (I would go with these), this code will also work fine.
#include <string.h>
#include <stdio.h>
int main(int argc, char** argv) {
char _p[] = "1000,Hey,There";
char* str1 = strtok(_p, ",");
char* str2 = strtok(NULL, "");
return 0;
}

Find longest word + length from string input

This is only my third or fourth program so bear with me:
I have to write a program that reads string input until EOF and then prints the longest word + its length. I'm not supposed to use strlen(). This is the skeleton code that's been provided -
#include <stdio.h>
static const int max_word_len = 50;
int main(int argc, const char *argv[]){
int c = 0;
int i = 0;
int longest;
char current[max_word_len];
char longest[max_word_len];
while(... != EOF && i < max_word_len-1){
}
}
I then have to make use of this:
void copy(char src[], char dst[], int count);
I've no clue how to proceed so any help is appreciated.

How do I concatenate strings via command line arguments separated by a +?

What I want to do is write arguments in the command line separated by a + and concatenate the arguments into a single string
eg:
./concat Wow + this + is + cool
Wow this is cool
I looked up a question for this sort of topic before but that involved concatenating only the first character of each argument and not the entire arguments. And it didn't ignore the separator
This is what I have
void concat(char **argv, int argc, char *string)
{
size_t i = 0;
for(int j=1; j<argc; j++)
{
string[i++] = *argv[j];
if(j+1 != argc)
{
string[i++] = ',';
string[i++] = ' ';
}
}
string[i] = '\0';
}
And this is what I'm doing in main to call this function
int main(int argc, char *argv[])
{
int allnum=0;
char string[1000];
concat(argv, argc, string);
printf("%s\n", string);
}
Using the strcpy or strcat in string.h is more simple to concatenate string.
For example:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(int argc, char const *argv[])
{
char * s = malloc (2 * argc *sizeof (char));
if (argc < 2)
exit(-1);
for(int i = 1; i < argc; i += 2) {
strcat(s, argv[i]);
strcat(s, " ");
}
printf("%s\n", s);
return 0;
}

Iterate through char array and print chars

I am trying to print each char in a variable.
I can print the ANSI char number by changing to this printf("Value: %d\n", d[i]); but am failing to actually print the string character itself.
What I am doing wrong here?
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
int len = strlen(argv[1]);
char *d = malloc (strlen(argv[1])+1);
strcpy(d,argv[1]);
int i;
for(i=0;i<len;i++){
printf("Value: %s\n", (char)d[i]);
}
return 0;
}
You should use %c format to print characters in C. You are using %s, which requires to use pointer to the string, but in your case you are providing integer instead of pointer.
The below will work. You pass in the pointer to a string when using the token %s in printf.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
int len = strlen(argv[1]);
char *d = malloc (strlen(argv[1])+1);
strcpy(d,argv[1]);
printf("Value: %s\n", d);
return 0;
}

Finding last character

I'm new to C and trying to create a function that check a string and returns the last character.
I get the function to print the correct letter, but I cant figure out how to return it :/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char last_chr(char *c);
int main (int argc, const char * argv[]) {
char *text[15];
strcpy(text, "*find:last;char#");
last_chr(text); //debugging
//printf("last char: %c", last_chr(text)); //not working
return 0;
}
char last_chr(char *c) {
char *endchr;
char result;
int pos = strlen(c)-1;
endchr = c[pos];
//sprintf(result,"%s",endchr); //"EXEC_BAD_ACCESS"
putchar(endchr); //prints #
//putc(endchr, result); //"EXEC_BAD_ACCESS"
//printf(endchr); //"EXEC_BAD_ACCESS"
return result;
}
You don't assign result. You probably mean
result = c[pos];
instead of endchr = c[pos];
endchr is a character-pointer instead of a character.
#include <stdio.h>
#include <string.h>
char last_chr(char *c);
int main (int argc, const char * argv[]) {
char text[32];//char *text[15]
strcpy(text, "*find:last;char#");//length is 17
printf("last char: %c", last_chr(text));//#
return 0;
}
char last_chr(char *c) {
if(c == NULL || *c == '\0') return 0;
return c[strlen(c)-1];
}

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